Can we solve the functions describing the bend of a cable at rest fixed at two positions?
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
add a comment |
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
add a comment |
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
optimization physics classical-mechanics applications
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
edited Jan 4 at 17:06
mathreadler
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
asked Jan 4 at 14:39
mathreadlermathreadler
14.8k72160
14.8k72160
add a comment |
add a comment |
1 Answer
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First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
add a comment |
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
add a comment |
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
answered Jan 4 at 16:45
AndreiAndrei
11.5k21026
11.5k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
add a comment |
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
Jan 4 at 17:01
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
Jan 4 at 17:10
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
Jan 4 at 17:42
add a comment |
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